You are on the right track!
x1=1,
x2=-1.455,
x3=-3.787,
x4=-2.140,
...
eventually it will settle on x=-2.28.
Remember that in the case of multiple roots, the one obtained by Newton's method is very dependent on the initial approximation. I assumed x1=1 is in radians.
Use Newton's method to approximate a root of the equation 3sin(x)=x as follows.
Let x1=1 be the initial approximation.
The second approximation x2 is
and the third approximation x3 is
I got x2=-1.454
but can't get x3 :(
3 answers
I agree with drwls's setup in
Newton's method
starting with x1=1 I also got
x2= -41.454 but then I got the next values as
x3 = -3.787
x4 = -2.1404
x5 = -2.2878
x6 = -2.1813
x7 = -2.283
etc
appears to converge to around x = -2.28
check:
3sin(-2.28) = -2.2766
Newton's method
starting with x1=1 I also got
x2= -41.454 but then I got the next values as
x3 = -3.787
x4 = -2.1404
x5 = -2.2878
x6 = -2.1813
x7 = -2.283
etc
appears to converge to around x = -2.28
check:
3sin(-2.28) = -2.2766
there was supposed to be a reference to
drwls setup in
http://www.jiskha.com/display.cgi?id=1283845535
As you can see in MathMate's reply both of our results coincide.
drwls setup in
http://www.jiskha.com/display.cgi?id=1283845535
As you can see in MathMate's reply both of our results coincide.